Edexcel AS/A2 Mathematics M1 - 8th June 2016 - Official Thread

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Does anyone have the Edexcel January 2016 IAL paper? Would really appreciate it if you could send me a link

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Reply 281

Zacken

Original post by dodgewc25

Anyone got any advice on how to do those "find the force exerted on the pulley by the string" questions? They're the only ones which stump me...

THanks in advance

Watch this https://www.youtube.com/watch?v=rC2IQ3R5im4

Reply 282

KINGYusuf

Can anyone help me with this Q? For Q I thought that time = t+1, however the mark scheme says t-1?

I don't understand :/

M1 1.png22.9KB

Reply 283

XxKingSniprxX

M1 June 2012 --> 66/75 = 80% = A. (I should have scored 71/75 raw marks) :sigh:

Mistakes & tips for future past papers:
- Understand how to resolve if a direct force is acting into the plane & split it up into horizontal/vertical components.
- Sub the value of T back into the vector position equation to make comparisons.
- Draw a SUVAT diagram carefully & understand when the displacement would be negative.

I'll be recording every past paper score that I do and my overall notes/opinion on that past paper.

Past papers completed = 2/13. (edexcel regular + edexcel IAL) [excluding x2 MAM pp closer to exam date]

Reply 284

nicoledsk

When do you know when to make u and s negative when using suvat equations? i.e in what context?

Reply 285

Don Pedro K.

Original post by nicoledsk

When do you know when to make u and s negative when using suvat equations? i.e in what context?

Well let's say you have a particle being projected upwards with a speed of u ms^-1 5 metres above the ground. If you were then to do a calculation which involves you working out at what speed the particle's travelling when the particle hits the ground, you COULD take u as being -u ms^-1 if you choose the DOWNWARDS direction to be positive. In this case, u would be negative because u ms^-1 is the speed when the particle is being projected up. Does that make sense?

Similarly for displacement in the scenario above, displacement would be -5 metres since the particle is ending up 5m below its starting position (imagine the particle going up from the point it was projected from, coming back down, going past this point, and then hitting the ground 5 metres below. The total displacement = -5m as a result).

Hope that explained it :smile:

Reply 286

fpmaniac

Can anyone help me with a few questions from the M1 edexcel book please:
Page 152: questions 6 and 7 ( i dont understand those type of questions in general)
Page 113: Question 16? I dont understand how the horizontal force works
Page 78 : Question 26c
Page 51: Example 15c
Also, the topic where there are 2 forces acting on a particle (plus its weight), I find it difficult to draw the triangle and use cosine rule. I can do it through resolving (e.g. Tsinx , Tcosx) but that makes it much more time consuming. So anyone got any tips for that.
Thats it for the timebeing. Thanks

(edited 7 years ago)

Reply 287

nicoledsk

Original post by Don Pedro K.

Thank you so much! it helps so much when you can actually understand the principles being applied to equations, thanks once again!

Reply 288

Don Pedro K.

Original post by nicoledsk

Thank you so much! it helps so much when you can actually understand the principles being applied to equations, thanks once again!

It certainly does haha :smile:

! No problem ^_^

Reply 289

fpmaniac

Original post by Zacken

Watch this https://www.youtube.com/watch?v=rC2IQ3R5im4

Hey, Can you or anyone explain to me how the diagrams work on pages 58 and 59 from the m1 book. Since the two blocks on top of each other are inside the lift, how would u draw the forces on each box. The book isnt very clear on what force represents what.
Thanks.

Reply 290

JP0458B

Is 3 papers a good amount for today? Or should I do another ?

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Reply 291

n2697

I NEED HELP WITH QUESTION 6 :https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/January%202015%20(IAL)%20QP%20-%20M1%20Edexcel.pdf
It says to find the greatest possible value of x - and in order to do that you take moments about B and assume the reaction force at A is 0 - but WHY?!

Reply 292

Ash1498

If the time that P has spent past A is t, Q has spent 1 less second than P past A (as it passed A 1 second later than P) so it's time is t-1.

Original post by KINGYusuf

Can anyone help me with this Q? For Q I thought that time = t+1, however the mark scheme says t-1?

I don't understand :/

Reply 293

JP0458B

3 papers enough to call it a day guys?

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Reply 294

Marxist

Original post by Don Joiner

3 papers enough to call it a day guys?

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It depends on the person to be honest.

Reply 295

Zacken

Original post by Don Joiner

3 papers enough to call it a day guys?

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lol srs? I haven't even done one paper in the past three days. :rofl:

Reply 296

Marxist

Original post by Zacken

lol srs? I haven't even done one paper in the past three days. :rofl:

Reply 297

KloppOClock

Original post by n2697

Imagine if the particle was all the way at the end at point C. The forces would be unbalanced and the rod would tip around the point B. Now imagine the particle is in the centre, the rod would remain in equilibrium. This means that the further away X is, the lower the reaction at A is. So what the question is asking you is to find the furthest value of X before the rod tips over. Now in order to this, you need to know this fact.

When X is the furthest away it can be before the rod tips, the value of the reaction at A must be 0.

You can then use that to take moments about B and solve the question.

Here is a more clearer explanation: